Driven Harmonic Motion and Resonance

(Topic)

Driven Harmonic Motion and Resonance

A free oscillator evolves from its initial state.

A driven oscillator also receives energy from an external time-dependent force.

The standard linear model is

|--------------------------| -m-¨x-+-b˙x-+-kx-=-F0-cosωt.-|

(1)

Here:

m is the mass,
b is the viscous damping coefficient,
k is the spring constant,
F₀ is the driving-force amplitude,
ω is the driving angular frequency.

Define

|----------| | ∘ k | ω0 = --,| --------m---

(2)

and

|--------| | -b--| γ = 2m .| ----------

(3)

Then

|--------------------------| | 2 F0- | ¨x + 2γ ˙x + ω0x = m cosωt.| ----------------------------

(4)

The solution contains two physically distinct parts:

|--------------------| x-(t)-=-xtr(t)-+-xss(t).-

(5)

The transient

xtr

(6)

depends on initial conditions and decays when

b > 0.

(7)

The steady-state response

xss

(8)

is determined by the driving force and remains after the transient has disappeared.

Figure 1. A driven damped oscillator contains an elastic restoring force, a viscous damping force, and an externally applied periodic force F₀ cosωt.

1 Homogeneous and particular solutions

The differential equation is linear:

m ¨x + b˙x + kx = F0 cosωt.

(9)

Its general solution is

|------------| |x = xh + xp,| --------------

(10)

where

m ¨xh + bx˙h + kxh = 0

(11)

and

m x¨p + bx˙p + kxp = F0cos ωt.

(12)

The homogeneous solution is exactly the damped free motion developed in M06-02.

For an underdamped system,

x = e−γt[C cosω t + D sin ω t], h d d

(13)

with

∘ -------- 2 2 ωd = ω 0 − γ .

(14)

Because this part decays, the long-time response is governed by the particular solution.

2 Why the steady state oscillates at the driving frequency

The external force is sinusoidal at angular frequency

ω.

(15)

For a linear time-invariant oscillator, a sinusoidal input produces a sinusoidal steady-state response at the same frequency.

The amplitude and phase can change, but the steady-state frequency is the driving frequency.

Write

|----------------------| |xss(t) = X cos(ωt − δ).| ------------------------

(16)

The response amplitude

X

(17)

and phase lag

δ

(18)

depend on

ω.

(19)

3 Direct sine-cosine derivation

Assume

xp = acosωt + csin ωt.

(20)

Then

˙xp = − aω sin ωt + cω cosωt,

(21)

and

¨xp = − aω2 cosωt − cω2 sinωt.

(22)

Substitute into

m ¨x + b˙x + kx = F0 cosωt.

(23)

The cosine coefficient gives

(k − m ω2)a + bωc = F . 0

(24)

The sine coefficient gives

2 (k − m ω )c − bωa = 0.

(25)

Solving,

|------------------2-----| |a = ---F0-(k-−-m-ω--)---,| -----(k-−-m-ω2)2 +-(bω)2--

(26)

and

|------------------------| |c = -------F0bω--------.| | (k − m ω2)2 + (bω )2 | --------------------------

(27)

4 Steady-state amplitude

Because

X2 = a2 + c2,

(28)

|------------------------------| | ---------F0-----------| X (ω ) = ∘ (k-−-m-ω2)2-+-(bω)2.| --------------------------------

(29)

Dividing numerator and denominator by

m,

(30)

|--------------------------------| | F0∕m | |X (ω) = ∘---2----2-2--------2-.| ------------(ω0-−-ω-)--+-(2γω-)--|

(31)

This is the frequency-response amplitude of the driven oscillator.

5 Steady-state phase lag

From

a = X cos δ, c = X sin δ,

(32)

tan δ = ---bω----. k − m ω2

(33)

The quadrant-aware form is

|-------------(------------)-| δ(ω ) = atan2 bω, k − m ω2 .| ------------------------------

(34)

Equivalently,

|-------------(-------------)--| |δ(ω) = atan2 2γ ω, ω20 − ω2 .| -------------------------------

(35)

The displacement lags the driving force by

δ.

(36)

6 Complex-exponential derivation

The same result follows more compactly with complex notation.

Represent the force as the real part of

F0eiωt.

(37)

Seek

z(t) = X^ei ωt.

(38)

Then

˙z = iωz,

(39)

and

¨z = − ω2z.

(40)

Substitution gives

( ) k − m ω2 + ibω X^ = F0.

(41)

Therefore

|--------------------| | F0 | X^ = -------2------.| ------k −-m-ω--+-ibω--

(42)

The physical response is

[ ] xss(t) = Re X^ei ωt .

(43)

The magnitude of

^ X

(44)

X,

(45)

and its argument is

− δ.

(46)

Figure 3. In the complex response denominator, k − mω² is the in-phase stiffness-inertia term and bω is the quadrature damping term. Their vector magnitude determines the inverse response amplitude.

7 Transient plus steady state

For an underdamped oscillator,

|--------−γt---------------------------------------| -x(t)-=-e---[C-cosωdt-+-D--sin-ωdt] +-X-cos(ωt-−--δ).-|

(47)

The constants

C

(48)

and

D

(49)

are determined by the initial state.

At long times,

−γt e → 0.

(50)

Thus

|-----------------------| x-(t)-−→--X-cos(ωt-−-δ).--

(51)

The system loses memory of its initial state but not of the continuing drive.

Figure 2. The total response is the sum of a decaying transient and a persistent steady-state oscillation at the driving frequency.

8 Normalized frequency response

Define the static displacement under force

F0

(52)

|------------| | F0-| Xstatic = k .| --------------

(53)

Define

ω r = --- ω0

(54)

and the damping ratio

ζ = -γ-. ω0

(55)

Then

|------------------------------| | X 1 | |------ = ∘-------2-2--------2.| -Xstatic-----(1-−-r-)-+--(2ζr)---

(56)

This dimensionless form shows how the response depends only on frequency ratio and damping ratio.

9 Low-frequency limit

ω ≪ ω0,

(57)

then

m ω2 ≪ k

(58)

and

bω ≪ k.

(59)

Therefore

|--------| X ≈ F0.| ------k---

(60)

Also,

δ ≈ 0.

(61)

The oscillator follows the slowly changing force almost quasistatically.

10 High-frequency limit

ω ≫ ω , 0

(62)

the inertial term dominates:

|k − m ω2 | ≈ m ω2.

(63)

Thus

|----------| X ≈ -F0-.| ------m-ω2--

(64)

The displacement becomes small, and

δ → π.

(65)

The displacement is nearly opposite the applied force.

11 At the undamped natural frequency

Set

ω = ω . 0

(66)

Then

k − m ω20 = 0.

(67)

Therefore

|--------------| | F0 | |X (ω0) = ----.| ----------bω0--

(68)

The phase lag is

|----------| δ(ω0 ) = π.| ---------2--

(69)

ω0,

(70)

the spring and inertial response terms cancel in the steady-state force balance, leaving damping to balance the applied force.

12 Displacement-amplitude resonance

The displacement amplitude is

--------F0∕m--------- X = ∘ --2----2-2-----2-2-. (ω0 − ω ) + 4γ ω

(71)

To maximize

X,

(72)

minimize the squared denominator:

2 22 2 2 D (ω) = (ω0 − ω ) + 4γ ω .

(73)

Differentiate:

	= −4ω(ω₀² − ω²) + 8γ²ω	(74)
	= 4ω.	(75)

For a nonzero resonance frequency,

|-----------------| | ∘ -2------- | ωr = ω0 − 2γ2. | ------------------

(76)

A nonzero displacement-amplitude peak exists only if

|--------| γ < ω√0-.| | 2 | ----------

(77)

Equivalently,

|--------| | 1 | ζ < √--.| -------2--

(78)

13 Maximum displacement response

ω = ω , r

(79)

ω2 = ω2 − 2γ2. r 0

(80)

The normalized maximum displacement is

|--------------------| |Xmax- = ---∘-1-----.| |F0∕k 2ζ 1 − ζ2 | ----------------------

(81)

For very light damping,

ζ ≪ 1,

(82)

ωr ≈ ω0

(83)

and

Xmax-≈ -1-≈ Q. F0∕k 2ζ

(84)

Figure 4. The normalized displacement response develops a tall narrow resonance peak for weak damping. Increased damping lowers and broadens the peak, and sufficiently strong damping removes the nonzero-frequency displacement maximum.

14 Natural frequency is not always the displacement-resonance frequency

Three frequencies should be distinguished:

undamped natural frequency:
$∘ ----- ω0 = k∕m,$ (85)
damped free-oscillation frequency:
$∘ -------- ωd = ω20 − γ2,$ (86)
driven displacement-resonance frequency:
$∘ --------- ωr = ω2 − 2γ2. 0$ (87)

For weak damping they are close, but they are not identical.

15 Phase response

The phase lag is

( ) δ = atan2 2γω, ω20 − ω2 .

(88)

Therefore:

δ → 0 as ω → 0,
δ = π∕2 at ω = ω₀,
δ → π as ω →∞.

The transition is sharper for weak damping.

Figure 5. The displacement response changes from nearly in phase with the force at low frequency to nearly opposite the force at high frequency, passing through a 90^∘ lag at ω₀.

16 Velocity response

The steady-state velocity is

vss = − ωX sin(ωt − δ).

(89)

Its amplitude is

|----------------| |V (ω) = ωX (ω ).| -----------------

(90)

ω = ω0,

(91)

|----| F0-- |F0- | V (ω0) = ω0 bω = | b .| 0 -----

(92)

The velocity-amplitude response is maximized at

--------- |ω = ω0. | ---------|

(93)

Thus velocity resonance occurs exactly at the undamped natural frequency for viscous damping.

17 Average power delivered by the drive

The applied force is

F (t) = F0cos ωt.

(94)

The velocity is

v(t) = − ωX sin(ωt − δ).

(95)

The instantaneous input power is

P(t) = F (t)v(t).

(96)

Averaging over one cycle gives

|--------------------| | 1 | |⟨P ⟩ = -F0 ωX sin δ.| --------2------------

(97)

Using the amplitude and phase formulas,

|----------------| |⟨P ⟩ = 1bω2X2. | --------2--------|

(98)

Equivalently,

|------------------------------| | 2 2 | |⟨P⟩ = F0--------bω----------.| --------2--(k −-m-ω2)2 +-(bω-)2

(99)

In steady state, average input power equals average power dissipated by damping.

18 Power resonance

The average power is maximized at

|--------| -ω-=-ω0.-|

(100)

At this frequency,

|------------2-| |⟨P ⟩max = F0-.| -----------2b--|

(101)

This is another reason to distinguish power resonance from displacement-amplitude resonance.

For finite damping:

ωr < ω0,

(102)

while the maximum average power occurs at

ω0.

(103)

19 Half-power frequencies

Define

ω −

(104)

and

ω+

(105)

1 ⟨P ⟩ = 2⟨P ⟩max.

(106)

For viscous damping, the condition becomes

|--------2-------| -|k-−-m-ω--| =-bω.|

(107)

The positive roots are

|-----∘--------------| |ω− = ω20 + γ2 − γ,| ----------------------

(108)

and

|-----∘--------------| | 2 2 | -ω+-=----ω0-+-γ-+--γ.-

(109)

Their difference is exactly

|--------------------------| | b- | |Δω = ω+ − ω − = 2γ = m .| ----------------------------

(110)

Their product is

|----------2-| -ω−-ω+-=-ω-0.|

(111)

Figure 6. The average absorbed power peaks at ω₀. The two half-power frequencies define the bandwidth Δω = ω₊ − ω₋ = 2γ.

20 Quality factor from resonance bandwidth

Define

|----------| | ω0 | |Q = Δ-ω-.| -----------

(112)

Since

Δ ω = 2γ,

(113)

------------------ | ω0 m ω0 | |Q = ---= -----.| ------2γ-----b----

(114)

Using

ζ = γ∕ω0,

(115)

|--------| | 1 | |Q = --.| ------2ζ--

(116)

For a lightly damped resonator, large

Q

(117)

means:

a tall displacement-response peak,
a narrow power-resonance bandwidth,
slow decay of free oscillations.

21 Example 1: steady-state amplitude and phase

Let

m	= 1.0 kg,	(118)
k	= 100 N∕m,	(119)
b	= 2.0 kg∕s,	(120)
F₀	= 10 N,	(121)
ω	= 8.0 rad∕s.	(122)

Then

ω0 = 10 rad∕s.

(123)

The in-phase denominator term is

k − m ω2 = 100 − 64 = 36 N∕m.

(124)

The damping term is

bω = 16 N ∕m.

(125)

Thus

X	=	(126)
	≈ 0.254 m.	(127)

The phase lag is

δ	= atan2(16, 36)	(128)
	≈ 0.418 rad	(129)
	≈ 23.96^∘.	(130)

22 Example 2: resonance frequency and quality factor

Use the same

m = 1.0 kg, k = 100N ∕m, b = 2.0 kg∕s.

(131)

Then

γ = 2-= 1.0 s−1. 2

(132)

The displacement-resonance frequency is

ω_r	=	(133)
	=	(134)
	≈ 9.90 rad∕s.	(135)

The power-resonance frequency is

|----------------| |ω0 = 10.0 rad ∕s.| -----------------

(136)

The bandwidth is

|---------------------| Δ ω = 2γ = 2.0rad∕s. | -----------------------

(137)

Therefore

|------------| | 10 | Q = ---= 5.| ------2-------

(138)

23 Mechanical impedance

The complex displacement relation is

( 2 ) ^ F0 = k − m ω + ibω X.

(139)

The complex velocity amplitude is

^V = iω ^X.

(140)

Define mechanical impedance by

|--------| | ^F | |Z = --.| ------^V--|

(141)

Then

Z	=	(142)
	= b + i.	(143)

The real part

b

(144)

represents dissipation.

The imaginary part represents reactive storage by mass and spring.

ω = ω0,

(145)

k m ω0 − ---= 0, ω0

(146)

|----------| Z-(ω0)-=-b.-

(147)

The impedance is purely resistive at power resonance.

24 Undamped driven oscillator

Set

b = 0.

(148)

Then

|--------------------| m ¨x + kx = F0 cosωt.| ----------------------

(149)

For

ω ⁄= ω0,

(150)

a particular solution is

|--------------------| | F0 ∕m | |xp = --2----2 cos ωt. ------ω0-−-ω----------

(151)

The denominator becomes small when the drive approaches the natural frequency.

Without damping, however, the behavior at exact resonance must be treated separately.

25 Beating for an undamped oscillator released from rest

Suppose

x (0 ) = 0, v(0) = 0,

(152)

and

ω ⁄= ω0.

(153)

The exact solution is

|--------------------------------| | F0∕m | x (t) = --2----2 [cos ωt − cosω0t].| -------ω-0 −-ω--------------------

(154)

Use

( ) ( ) cosa − cos b = − 2sin a-+-b sin a-−-b . 2 2

(155)

Then

|--------------------(---------)----(--------)---| |x(t) = − 2F0-∕m--sin ω-+-ω0-t sin ω-−-ω0-t . | | ω20 − ω2 2 2 | -------------------------------------------------

(156)

When

ω

(157)

is close to

ω0,

(158)

the rapid oscillation is modulated by a slow envelope.

This is beating.

Figure 7. In the undamped oscillator driven near but not exactly at its natural frequency, interference between the drive-frequency and natural-frequency terms produces beats.

26 Exact undamped resonance

Now set

ω = ω0.

(159)

The ordinary cosine trial particular solution fails because

cos ω0t

(160)

is already a homogeneous solution.

Try

x = Ct sinω t. p 0

(161)

Substitution gives

F0- 2C ω0cos ω0t = m cos ω0t.

(162)

Therefore

|------------| |C = --F0--.| ------2m-ω0--|

(163)

Hence

|----------------------| |xp(t) = -F0--t sin ω0t.| ---------2m-ω0---------|

(164)

The oscillation amplitude grows linearly with time.

The ideal undamped resonance is therefore not a finite sinusoid of infinite amplitude.

It is a secularly growing response.

Figure 8. At exact resonance with zero damping, the response contains a factor of t. The envelope grows linearly rather than approaching a finite steady-state amplitude.

27 Why damping regularizes resonance

With

b > 0,

(165)

energy is continuously removed at rate

2 Pd = − bv .

(166)

In steady state, the drive supplies exactly the average energy dissipated each cycle.

This balance produces a finite resonance response.

Smaller

b

(167)

means less dissipation per cycle and therefore a larger resonant response.

In the formal limit

b → 0,

(168)

the finite damped steady-state picture breaks down at exact resonance and the secular undamped solution must be used.

28 Resonance as energy transfer

Resonance is fundamentally efficient energy transfer from a periodic drive into an oscillator.

At low frequency, displacement follows the force nearly in phase and velocity is small.

At high frequency, inertia suppresses displacement.

Near

ω0,

(169)

the force is nearly in phase with velocity, so the drive performs positive net work over each cycle most effectively.

At exact power resonance,

δ = π, 2

(170)

so velocity is in phase with force.

That is why average absorbed power is maximized at

ω0.

(171)

29 Resonance is not always desirable

Resonance can be useful in:

frequency selection,
musical instruments,
sensors,
resonant cavities,
mechanical and electrical filters.

It can be dangerous in:

machinery,
structures,
vehicles,
rotating equipment,
components subjected to periodic forcing.

The linear oscillator provides the first model for both resonance enhancement and resonance avoidance.

30 Common mistakes

Assuming the steady-state response oscillates at the natural frequency rather than the driving frequency.
Forgetting that the total solution contains both transient and steady-state parts.
Using the undamped natural frequency ω₀ as the displacement-resonance frequency for finite damping without qualification.
Confusing ω_d, ω_r, and ω₀.
Using a one-argument inverse tangent for the phase and losing the correct quadrant.
Forgetting that displacement lags force by π∕2 at ω₀.
Saying resonance means an infinite steady-state amplitude when damping is nonzero.
Saying the undamped exact-resonance solution is an ordinary sinusoid with infinite amplitude.
Forgetting that exact undamped resonance produces a secular factor t.
Confusing displacement resonance with power resonance.
Assuming the half-power bandwidth is centered arithmetically on ω₀; instead ω₋ω₊ = ω₀².
Forgetting that the exact power bandwidth for viscous damping is Δω = 2γ.
Treating Q only as a decay parameter and missing its resonance-bandwidth interpretation.
Forgetting that the mechanical impedance is purely real at ω₀.
Applying the linear frequency-response formulas when stiffness, damping, or forcing is strongly nonlinear.

31 Practice exercises

Derive the steady-state amplitude by using a cosine-sine trial solution.
Derive the phase lag and explain why atan2 is preferable to a one-argument inverse tangent.
Repeat the steady-state derivation using complex exponentials.
Derive the normalized response X∕(F₀∕k) in terms of r = ω∕ω₀ and ζ.
Derive the low-frequency and high-frequency limits of the displacement response.
Show that δ = π∕2 at ω = ω₀.
Maximize the displacement amplitude to derive ω_r = .
Derive the condition ζ < 1∕ for a nonzero-frequency displacement peak.
Show that the velocity-amplitude response is maximized at ω₀.
Derive the average power formula and show that power is maximized at ω₀.
Derive the exact half-power frequencies and show that Δω = 2γ.
Show that ω₋ω₊ = ω₀².
Derive Q = ω₀∕(2γ) = mω₀∕b from the bandwidth.
Derive the beat solution for the undamped oscillator released from rest.
Derive the secular solution x_p = (F₀∕2mω₀)t sin ω₀t at exact undamped resonance.

32 Summary

The driven damped oscillator obeys

|--------------------------| -m-¨x-+-b˙x-+-kx-=-F0-cosωt.-|

(172)

After transients decay, the steady-state response is

|--------------------------| xss-=-X-(ω-)cos[ωt-−-δ(ω)].-

(173)

The amplitude is

|------------------------------| | F0 | X (ω ) = ∘---------2-2-------2.| -----------(k-−-m-ω-)--+-(bω)---

(174)

The phase lag is

|----------(------------)--| |δ = atan2 bω, k − m ω2 .| ---------------------------

(175)

The displacement-amplitude resonance occurs at

|-----∘-----------| | 2 2 | ωr-=----ω0-−-2γ-,-|

(176)

when that expression is real.

Power resonance occurs at

|--------| -ω-=-ω0.-|

(177)

The half-power bandwidth is

|------------b--| Δ ω = 2γ = --, | ------------m----

(178)

and

|----------------| |Q = ω0-= m-ω0-.| | 2γ b | ------------------

(179)

In the undamped limit, near-resonant forcing produces beats, while exact resonance produces secular growth:

|--------------------| |xp = --F0--tsinω0t. | ------2m--ω0---------|

(180)

M06-04 extends linear oscillation theory from one coordinate to multiple coupled coordinates and introduces normal modes.

References

[1] PhysicsLibrary, M06-02, Damped Harmonic Motion.

[2] PhysicsLibrary, M06-01, Simple Harmonic Motion.

[3] A. P. French, Vibrations and Waves, W. W. Norton, 1971.

[4] J. R. Taylor, Classical Mechanics, University Science Books, 2005.

"Driven Harmonic Motion and Resonance" is owned by bloftin.
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Driven Harmonic Motion and Resonance